Tính: \(H=cos\dfrac{\pi}{19}+cos\dfrac{3\pi}{19}+...+cos\dfrac{17\pi}{19}\)
Tính: \(H=cos\dfrac{\pi}{19}+cos\dfrac{3\pi}{19}+...+cos\dfrac{17\pi}{19}\)
\(H.sin\dfrac{\pi}{19}=sin\dfrac{\pi}{19}.cos\dfrac{\pi}{19}+sin\dfrac{\pi}{19}cos\dfrac{3\pi}{19}+...+sin\dfrac{\pi}{19}cos\dfrac{17\pi}{19}\)
\(=\dfrac{1}{2}sin\dfrac{2\pi}{19}+\dfrac{1}{2}sin\dfrac{4\pi}{19}-\dfrac{1}{2}sin\dfrac{2\pi}{19}+...+\dfrac{1}{2}sin\dfrac{18\pi}{19}-\dfrac{1}{2}sin\dfrac{16\pi}{19}\)
\(=\dfrac{1}{2}sin\dfrac{18\pi}{19}=\dfrac{1}{2}sin\left(\pi-\dfrac{\pi}{19}\right)=\dfrac{1}{2}sin\dfrac{\pi}{19}\)
\(\Rightarrow H=\dfrac{1}{2}\)
Hãy tính \(\sin\alpha\) và \(tg\alpha\) nếu :
a) \(\cos\alpha=\dfrac{5}{13}\)
b) \(\cos\alpha=\dfrac{15}{17}\)
c) \(\cos\alpha=0,6\)
a: \(\sin a=\sqrt{1-\left(\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\)
\(\tan a=\dfrac{12}{5}\)
b: \(\sin a=\sqrt{1-\left(\dfrac{15}{17}\right)^2}=\dfrac{8}{17}\)
\(\tan a=\dfrac{8}{15}\)
c: \(\sin a=\sqrt{1-0.6^2}=0.8\)
nên \(\tan a=\dfrac{4}{3}\)
Bài 1 : thực hiện các phép tính ( tính hợp lí nếu có thể ):
a ,\(\dfrac{7}{2}.\dfrac{8}{13}+\dfrac{8}{13}.\dfrac{-5}{2}+\dfrac{8}{13}\)
b , \(\dfrac{-5}{17}.\dfrac{-9}{23}+\dfrac{9}{23}.\dfrac{-22}{17}+11\dfrac{9}{23}\)
Bài 2 : tìm x , biết :
a , \(\dfrac{7}{8}+x=\dfrac{3}{5}\)
b , \(\dfrac{1}{3}:\left(2x-1\right)=\dfrac{-4}{24}\)
Bài 2 :
a, \(x=\dfrac{3}{5}-\dfrac{7}{8}=\dfrac{24-30}{40}=-\dfrac{6}{40}=-\dfrac{3}{20}\)
b, \(2x-1=-2\Leftrightarrow x=-\dfrac{1}{2}\)
Tính giá trị của biểu thức :
\(A=\dfrac{2\cos^2\dfrac{\pi}{8}-1}{1+8\sin^2\dfrac{\pi}{8}\cos^2\dfrac{\pi}{8}}\)
\(\cos\dfrac{\pi}{4}=\cos2\left(\dfrac{\pi}{8}\right)=2\cos^2\dfrac{\pi}{8}-1=\dfrac{\sqrt{2}}{2}\)
\(8\sin^2\dfrac{\pi}{8}.\cos^2\dfrac{\pi}{8}=2\left(2\sin\dfrac{\pi}{8}.\cos\dfrac{\pi}{8}\right)^2=2.\sin^2\dfrac{\pi}{4}=1\)
Vậy A=\(\dfrac{\sqrt{2}}{4}\)
\(A=\dfrac{2\left(cos\dfrac{\Pi}{2}\right)^2-1}{1+8sin^2\dfrac{\Pi}{8}\left(cos\dfrac{\Pi}{2}\right)^2}=\dfrac{2\left(0\right)^2-1}{1+8sin^2\left(0\right)^2}=\dfrac{-1}{1}=-1\)
vậy A= -1
Câu này đề cũng sai nhưng mình sẽ giải theo kiểu đề sai, HY VỌNG NGƯỜI CHÉP ĐỀ LÊN CẨN THẬN HƠN KẺO MẤT THỜI GIAN ,CÔNG SỨC NGƯỜI GIẢI MÀ CHẲNG ĐƯỢC GÌ.
Cho \(\dfrac{\sin^4\alpha}{a}+\dfrac{\cos^4\alpha}{b}=\dfrac{1}{a+b}\).CM:\(\dfrac{\sin^8\alpha}{a^3}+\dfrac{cos^8\beta}{b^3}=\dfrac{1}{\left(a+b\right)^3}\)
Thực hiện phép tính( tính nhanh nếu có thể)
a, \(\left(-\dfrac{1}{2}\right)^2.\dfrac{7}{4}:\left(\dfrac{5}{8}-1\dfrac{3}{16}\right)\)
b, \(17\dfrac{6}{11}.\dfrac{4}{27}-8\dfrac{6}{11}:\dfrac{27}{4}+350\%\)
a) Ta có: \(\left(\dfrac{-1}{2}\right)^2\cdot\dfrac{7}{4}:\left(\dfrac{5}{8}-1\dfrac{3}{16}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{7}{4}:\left(\dfrac{5}{8}-\dfrac{19}{16}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{7}{4}:\dfrac{-9}{16}\)
\(=\dfrac{1}{4}\cdot\dfrac{7}{4}\cdot\dfrac{-16}{9}\)
\(=\dfrac{-112}{144}=\dfrac{-7}{9}\)
b) Ta có: \(17\dfrac{6}{11}\cdot\dfrac{4}{27}-8\dfrac{6}{11}:\dfrac{27}{4}+350\%\)
\(=17\dfrac{6}{11}\cdot\dfrac{4}{27}-8\dfrac{6}{11}\cdot\dfrac{4}{27}+350\%\)
\(=\dfrac{4}{27}\left(17+\dfrac{6}{11}-8-\dfrac{6}{11}\right)+\dfrac{7}{2}\)
\(=\dfrac{4}{27}\cdot9+\dfrac{7}{2}\)
\(=\dfrac{4}{3}+\dfrac{7}{2}=\dfrac{8}{6}+\dfrac{21}{6}=\dfrac{29}{6}\)
hãy tính sin a và tan a nếu
a,cos a = 5/13
b, cos a = 15/17
a) cos a = 5/13 <=> a=670 23'
+ sin a=0,9
+ tan a = 2,4
b) cos a= 15/17 <=> a=28o 4'
+ sin a =0,5
+ tan a= 0,5
Giải các phương trình lượng giác:
a) sin8x + cos8x = \(\dfrac{17}{16}\)cos22x
b) sin2x + sin22x + sin23x = 2
c) 2cos22x + cos2x = 4 sin22xcos2x
d) 2cos6x + tan3x = \(\dfrac{4}{5}\)
a. cho sin = 8/17 . Tính cos , tan , cot
b. cho cot = 3/4 . Tính cos , sin , cot
Lớp 9 nên coi như các góc này đều nhọn
a.
\(cosa=\sqrt{1-sin^2a}=\dfrac{15}{17}\)
\(tana=\dfrac{sina}{cosa}=\dfrac{8}{15}\)
\(cota=\dfrac{1}{tana}=\dfrac{15}{8}\)
b.
\(1+cot^2a=\dfrac{1}{sin^2a}\Rightarrow sina=\dfrac{1}{\sqrt{1+cot^2a}}=\dfrac{4}{5}\)
\(cosa=\sqrt{1-sin^2a}=\dfrac{3}{5}\)
\(tana=\dfrac{1}{cota}=\dfrac{4}{3}\)
a) \(\cos=\sqrt{1-\sin^2}=\sqrt{1-\dfrac{64}{289}}=\dfrac{15}{17}\)
\(\tan=\dfrac{\sin}{\cos}=\dfrac{8}{17}:\dfrac{15}{17}=\dfrac{8}{15}\)
\(\cot=\dfrac{\cos}{\sin}=\dfrac{15}{17}:\dfrac{8}{17}=\dfrac{15}{8}\)
Giải PT
a1) \(3.\cos4x-2^{ }\cos^23x=1\)
a2) \(2\cos2x-8\cos x+7=\dfrac{1}{\cos x}\)
a3) \(\dfrac{\left(1+\sin x+\cos2x\right)\sin\left(x+\dfrac{\pi}{4}\right)}{1+\tan x}=\dfrac{1}{\sqrt{2}}\cos x\)
a4) \(9\sin x+6\cos x-3\sin2x+\cos2x=8\)
a) Pt \(\Leftrightarrow3.cos4x-\left(cos6x+1\right)=1\)
\(\Leftrightarrow3cos4x-cos6x-2=0\)
Đặt \(t=2x\)
Pttt:\(3cos2t-cos3t-2=0\)
\(\Leftrightarrow3\left(2cos^2t-1\right)-\left(4cos^3t-3cost\right)-2=0\)
\(\Leftrightarrow-4cos^3t+6cos^2t+3cost-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cost=1\\cost=\dfrac{1+\sqrt{21}}{4}\left(vn\right)\\cost=\dfrac{1-\sqrt{21}}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}t=k2\pi\\t=\pm arc.cos\left(\dfrac{1-\sqrt{21}}{4}\right)+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\dfrac{1}{2}.arccos\left(\dfrac{1-\sqrt{21}}{4}\right)+k\pi\end{matrix}\right.\) (\(k\in Z\))
Vậy...
a2) \(2cos2x-8cosx+7=\dfrac{1}{cosx}\) (ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\))
\(\Leftrightarrow2.\left(2cos^2x-1\right)-8cosx+7=\dfrac{1}{cosx}\)
\(\Leftrightarrow2.\left(2cos^2x-1\right)cosx-8cos^2x+7cosx=1\)
\(\Leftrightarrow4cos^3x-8cos^2x+5cosx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\) (tm) (\(k\in Z\))
Vậy...
a3) Đk: \(x\ne-\dfrac{\pi}{4}+k\pi;x\ne\dfrac{\pi}{2}+k\pi\)
Pt \(\Leftrightarrow\dfrac{\left(1+sinx+1-2sin^2x\right).\dfrac{1}{\sqrt{2}}\left(sinx+cosx\right)}{1+\dfrac{sinx}{cosx}}=\dfrac{1}{\sqrt{2}}cosx\)
\(\Leftrightarrow\dfrac{\left(-2sin^2x+sinx+2\right).\left(sinx+cosx\right)cosx}{cosx+sinx}=cosx\)
\(\Leftrightarrow\left(2+sinx-2sin^2x\right).cosx=cosx\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\2+sinx-2sin^2x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\) (\(k\in Z\))
Vậy...
a4) Pt \(\Leftrightarrow9sinx+6cosx-6sinx.cosx+1-2sin^2x=8\)
\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sin^2x-9sinx+7\right)=0\)
\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sinx-7\right)\left(sinx-1\right)=0\)
\(\Leftrightarrow\left(1-sinx\right)\left(6cosx+2sinx+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\6cosx+2sinx=7\left(vn\right)\end{matrix}\right.\) (\(6cosx+2sinx=7\) vô nghiệm do \(6^2+2^2< 7^2\))
\(\Rightarrow sinx=1\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi;k\in Z\)
Vậy...